Finite Elements Methods

Formulary for Prof. Estor�'s exam

Finite Element Method in General

One wants to obtain the equilibrium eqautions for the body, discretized by �niteelements in the form

M · U + C · U +K · U = R

Displacement of the nodes:

U =[U1 U2 . . . Un

]Tn: degrees of freedom

Displacement within the element m:

u(m)(x, y, z) = H(m)(x, y, z)U

H(m) : Displacement interpolation matrix

Strain inside the element m:

ε(m)(x, y, z) = B(m)(x, y, z)U

B(m) : Verzerrungs-Verschiebungs-Matrix

Play around with those matrices and from the priciple of virtual displacement´V

εT τ dV =´V

UT fBdV +´S

UST fS dS +∑i

U iTF i it follows:

System of equations (stationary case KU = R)[∑m

´v(m)

B(m)TC(m)B(m) dV (m)

]U =

∑m

´v(m)

H(m)T fB(m) dV (m)

+∑m

´S(m)

HS(m)T fS(m) dS(m) +∑m

´v(m)

B(m)T τ I(m) dV (m) + F

1

From that one can obtain a formular to calculate each of the matrices in theequlibrium equasions:

Sti�ness matrix: K =∑m

´V (m)

B(m)TC(m)B(m)dV (m) =∑K(m)

Volume forces: RB =∑m

´V (m)

H(m)T fB(m) dV (m) =∑mR

(m)B

Surface forces: RS =∑m

´S(m)

HS(m)T fS(m) dS(m) =∑mR

(m)S

Initial stresses: RI =∑m

´v(m)

B(m)T τ I(m) dV (m) =∑mR

(m)I

Single forces: RC = F

With d'Alembert Principle the matrices for the time dependent case follow:

Mass matrix: M =∑m

´V (m)

ρ(m)H(m)TH(m) dV (m) =∑mM (m)

Damping matrix: C =∑m

´V (m)

κ(m)H(m)TH(m) dV (m) =∑mC(m)

According to the choosen elements one has to concidere only certain componentsof stress, strain and displacement:

2

(Isoparametric) Truss Elements

Displacement of the nodes:

Global: U =[u1 v1 u2 v2 . . . vn

]TLocal: U =

[u1 v1 u2 v2 . . . vn

]Tn: number of nodes in element

U can be linked with the overall node displacement vector of the system (has thesame orientation).

Interpolation:

Coordinates: x(r) =∑hixi Displacements: u(r) =

∑hiui

In the following, an element with 2 nodes is considered.

Displacement interpolation matrix:

H =[h1 0 h2 0

]=[

12 (1− r) 0 1

2 (1 + r) 0]with

x(r) = H · Xu(r) = H · U

Setting up the strain displacement matrix:

εx = dudx

ε = B · U=⇒ B =

[h1x 0 h2x 0

]=[− 1

L 0 1L 0

]because hix = ∂hi

∂x = ∂hi

∂r∂r∂x = ∂hi

∂r

(∂x∂r

)−1= ∂hi

∂r

(L2

)−1

Setting up the sti�ness matrix:

K =´V

BTCB dV = AEL

0

BT B dx with A: Cross-sectional area of truss

It follows: K = EAL

1 0 −1 00 0 0 0−1 0 1 00 0 0 0

Transformed to global directions:

K = TT KT = EAL

cos2α cosα sinα −cosα cosβ −cosα sinβ

cosα sinα sin2α −sinα cosβ −sinα sinβ−cosα cosβ −sinα cosβ cos2β sinβ cosβ−cosα sinβ −sinα sinβ sinβ cosβ sin2β

3

4

Isoparametric Plate Elements

Displacement of the nodes:

U =[u1 . . . un v1 . . . vn

]Tn: number of nodes in element

U can be linked with the overall node displacement vector of the system.

Interpolation:

Coordinates: x(r, s) =∑hixi; y(r, s) =

∑hiyi

Displacements: u(r, s) =∑hiui; v(r, s) =

∑hivi

In the following formulars, an element with 4 nodes is concidered.

Displacement interpolation matrix:

H =

[h1 h2 h3 h4 0 0 0 00 0 0 0 h1 h2 h3 h4

]with

X(r, s) = H · XU(r, s) = H · U

Match H with the vector of displacements of all nodes U to obtain the globalsti�ness matric H(m) for the element.

Setting up the strain displacement matrix:

εx = dudx εy = dv

dy

εxy = dudy + dv

dx

ε = B · U=⇒ B =

h1x h2x h3x h4x 0 0 0 00 0 0 0 h1y h2y h3y h4y

h1y h2y h3y h4y h1x h2x h3x h4x

Match B with the vector of displacements of all nodes U to obtain the global sti�-ness matric B(m) for the element.

Calculating derivatives of hi:[hixhiy

]=

[ ∂hi

∂x∂hi

∂y

]= J−1 ·

[∂hi

∂r∂hi

∂s

]with J =

[∂x∂r

∂y∂r

∂x∂s

∂y∂s

]Inverse 4x4 matrix: A =

[a bc d

]−→ A−1 = 1

det(A)

[d −b−c a

]Setting up the sti�ness matrix:

K =´V

BTCB dV = t ·1

−1

1

−1

BTCB dr ds t: thickness of the element

Match K with the vector of displacements of all nodes U to obtain the globalsti�ness matric K(m) for the element.

5

6

Beam Elements

Displacement of the nodes:

U =[w1 ϕ1 w2 ϕ2

]Tfor an element with 2 nodes

Displacement interpolation matrix (Hermite Beam):

w(x) = HU

H =[

1− 3 x2

L2 + 2 x3

L3 x− 2x2

L + x3

L2 3 x2

L2 − 2 x3

L3 −x2

L + x3

L2

]

=⇒ K = EI

12L3

6L2

4L

− 12L3 − 6

L212L3

6L2

2L − 6

L24L

For di�erent H the strain displacement matrix B has to be determined �rst tocalculate K =

´V

BTCB dV .

Isoparametric:

Displacement of the nodes:

U =[w1 . . . wq ϕ1 . . . ϕq

]Tq: number of nodes in element

Interpolation:

w(r) =∑hiwi ϕ(r) =

∑hiϕi

Use the interpolation functions for a 1D element.

For displacement interpolation matrix H and strain interpolation matrix B con-cider displacement and rotation seperatly:

Hw =[h1 . . . hq 0 . . . 0

]Hϕ =

[0 . . . 0 h1 . . . hq

]Bw = J−1

[∂h1

∂r . . .∂hq

∂r 0 . . . 0]

Bϕ = J−1[

0 . . . 0 ∂h1

∂r . . .hq

∂r

]with the Jacobian J = ∂x

∂r

7

Material Matrices

8

Calculating Loads

In general RS =´S

HST q(x) dS for distributed loads q(x).

In the following a load acting between note 1 and 2 of a 4-node-isoparametricelement where s = const = 1 is considered.

Set up HS matrix

Consider only the displacements of the two nodes contained in the surface theload is working on:[uSvS

]= HSUS mit US =

[u1 u2 v1 v2

]HS =

[12 (1 + r) 1

2 (1− r) 0 00 0 1

2 (1 + r) 12 (1− r)

]Integrate

Rewrite q(x) in local coordinates to obtain q(r) and integrate over the surface:

R =[Rx1 Rx2 Ry1 Ry2

]T= t ·

1

−1

HST q(r) det(JS) dr

with q(r) =

[qx(r)qy(r)

]=fS and t: thickness of the plate

Jabobian matrix

Since s = const here, the Jacobian degenerates to JS =[

∂x∂r

∂y∂r

].

Because JS is not quadratic anymore, one has to use the Gramian determinant:

det(JS) =√JSJST =

√(∂x∂r

)2+(

∂y∂r

)2

Load in one direction (after the other)

For a vertical load with qx(x) = 0 only the vertical displacement and forces haveto be concidered:

vS = HS

[v1

v2

]=⇒ HS =

[h1 h2

]=[

12 (1 + r) 1

2 (1− r)]

R =

[Ry1

Ry2

]= t ·

1

−1

HST qy(r) det(J) dr

Always try set up single forces that representate the distributed load �rst!

9

Setting up Mass Matrices

M (m) =´

V (m)

ρ(m)H(m)TH(m) dV (m)

Consistent mass matrix:

1D: M (m) =L

0

A(x)ρ(m)H(m)TH(m) dx =1

−1

A(r)ρ(m)H(m)TH(m)det(J) dr

2D: M (m) =1

−1

1

−1

t(x)ρ(m)H(m)TH(m)det(J) dr ds

Lumped mass matrix:

M (m) =

mn 0 0

0. . . 0

0 0 mn

with m: Mass of the element; n: Number of

nodes in element

Calculating Stresses

In general: σ = C ·BU

Truss: σ = Cε = E∆LL with ∆L = u2 − u1

Plate (at s = s∗, r = r∗): σ = C ·B |r∗, s∗ U

10

Jacobian Matrix

Transformation of coordinates / chain rule /...:

J =

∂x∂r

∂y∂r

∂z∂r

∂x∂s

∂y∂s

∂z∂s

∂x∂t

∂y∂t

∂z∂t

so that

∂∂r∂∂s∂∂t

= J

∂∂x∂∂y∂∂z

Used to calculate: ∂h

∂x∂h∂y∂h∂z

= J−1

∂h∂r∂h∂s∂h∂t

and´V

. . . dV =1

−1

1

−1

1

−1

. . . det(J) dr ds dt

Rules for Variational Operator

δ dudx = d

dxδu δb

a

f(x) dx =b

a

δf(s) dx

Integration by Parts

b

a

u′(x)v(x) dx = [u(x)v(x)]ba −

b

a

u(x)v′(x) dx

Variation of the Total Potential

Total potential: Π = U −WU : Strain energy W : Potential of external loads

Obtain natural boundary conditions and partial di�erential equation by solvingδΠ = 0, where Π is the (hopefully given) potential of the system.

Ritz Method

Insert an approximation function for displacement with coe�cients a1, a2,a3, ...and solve ∂Π

∂a1, ∂Π

∂a2, ...

Note that ∂∂a1

b

a

f(ai) dx =b´

a

∂∂a1

f(ai) dx.

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Gauss Integration

K =´V

F (r, s, t) dr ds dt mit F (r, s, t) = BTCB det(J)

Gauss:b

a

F (r) dr = α1F (r1) + α2F (r2) + ...+ αnF (rn)

Polynomials up to the order (2n− 1) can be integrated axactly.

Example

The integral1

−1

F (r) dr shall be approximated using

just one approximation point:1

−1

F (r) dr = 2 · F (0)

two approximation points:1

−1

F (r) dr = 1 · F (−0.57735) + 1 · F (0.57735)

13

Newton-Cotes Integration

K =´V

F (r, s, t) dr ds dt mit F (r, s, t) = BTCB det(J)

Newton-Cotes:b

a

F (r) dr = (b− a)n∑

i=0

Cni Fi +Rn

with Cni : Newton-Cotes constants; Rn: Remainder (error estimation)

The approximation points Fi are linearly distributed, for the distance betweento points h = b−a

n holds.

Example

The integral1

−1

F (r) dr shall be approximated using

just one interval:1

−1

F (r) dr = 2 ·(

12F (−1) + 1

2F (1) +R1

)two intervals:1

−1

F (r) dr = 2 ·(

16F (−1) + 4

6F (0) + 16F (1) +R2

)

14

Centrale Di�erences Method

tU = 1∆t2

(t−∆tU − 2 tU +t+∆t U

)tU = 1

2∆t

(−t−∆tU +t+∆t U

)Error is of order (∆t)2.

System of equations at time t:

M tU + C tU +K tU =t R

Because the system is concidered at time t, not at time t+∆t the method is calledand explicit integration method.

Insert approximations for tU and tU :(1

∆t2M + 12∆tC

)t+∆tU =t R−

(K − 2

∆t2M)tU −

(1

∆t2M −1

2∆tC)t−∆tU

Since 0U , 0U and 0U are known, one only needs to calculate −∆tU to start thecalculation of one time step after another.

From the equations for tU and tU one obtains: −∆tU =0 U −∆t0U + ∆t2

20U

Now one can calculate ∆tU and respectivly t+∆tU and then tU and tU for everytime step.

15

Newmark Integration

Assumptions:

t+∆tU =t U +[(1− δ) tU + δt+∆tU

]∆t

t+∆tU =t U + tU∆t+[(

12 − α

)tU + αt+∆tU

]∆t2

Obtain t+∆tU from second equation, insert in �rst equation.

Insert the resulting eqations for t+∆tU and t+∆tU in the euation of motion attime t+ ∆t.

M t+∆tU + C t+∆tU +K t+∆tU =t+∆t R

Because the system is concidered at time t+∆t, not at time t the method is calledand implicit integration method.

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All �gures and tables are taken from the book �Finite-Element-

Procedures� by Klaus-Jürgen Bathe. It is the only source I used...

SoSe 11, Lars Radtke